Question

In a Mendelian dihybrid cross, out of 400 \(F_2\) generation plants, how many will be homozygous for at least one gene?

• A. 100
• B. 200
• C. 300
• D. 400

Hint:

In a dihybrid cross, only the double heterozygote \(AaBb\) (4 out of 16, or 1/4) is heterozygous for both genes.
All other 12 genotypes (12 out of 16, or 3/4) are homozygous for at least one gene.
\(400 \times 12/16 = 300\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a classical Mendelian dihybrid cross, the \(F_2\) generation is produced by selfing double heterozygous (\(AaBb\)) \(F_1\) individuals.
The two genes assort independently, producing predictable genotype and phenotype frequencies.
Key Formula or Approach:
An individual can be either homozygous or heterozygous for any given gene.
For a single gene locus, the probability of an \(F_2\) offspring being heterozygous (e.g., \(Aa\)) is \(1/2\), and the probability of being homozygous (\(AA\) or \(aa\)) is \(1/2\).
For two independently assorting genes, the probability of being heterozygous for BOTH genes is: \[ P(\text{heterozygous for both}) = P(\text{Het}_1) \times P(\text{Het}_2) \] The probability of being homozygous for at least one gene is the complement of being heterozygous for both: \[ P(\text{homozygous for at least one}) = 1 - P(\text{heterozygous for both}) \]

Step 2: Detailed Explanation:
Let us calculate the probability of being heterozygous for both genes:
For the first gene (A), the probability of heterozygosity (\(Aa\)) is \(1/2\).
For the second gene (B), the probability of heterozygosity (\(Bb\)) is \(1/2\).
Because the two genes assort independently, the probability of an \(F_2\) plant being heterozygous for both genes (\(AaBb\)) is: \[ P(\text{heterozygous for both}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \] The probability of being homozygous for at least one gene is: \[ P(\text{homozygous for at least one}) = 1 - \frac{1}{4} = \frac{3}{4} \] Out of 400 total \(F_2\) plants, the expected number of plants homozygous for at least one gene is: \[ \text{Expected number} = 400 \times \frac{3}{4} = 300 \text{ plants} \]

Step 3: Final Answer:
Therefore, the expected number of plants is 300, which corresponds to option (C).